00:01
For this problem on the topic of electrostatics, we want to find the potential at a distance z above the center of the three charge distributions that we are shown in the figure, and then in each case compute the electric field e as minus nabler v.
00:17
Now, we want to then suppose that the right -hand charge in a is changed to minus q and find the potential at p, and we want to know what this suggests.
00:27
Now, for the first charge distribution, we have the potential.
00:32
V equal to 1 over 4 pi epsilon not times the product of charges 2q or rather the total charge 2q over the square root of the distance z squared plus d over 2 all squared.
01:06
For b we have the potential v equal to 1 over 4 pi epsilon not times the integral from minus l to l of lambda d x divided by the square root of z squared plus x squared and this is lambda over 4 pi epsilon not times the natural log of x plus the square root of z squared plus x squared and this is evaluated from minus l to l.
01:54
And so we get the potential to be lambda over 4 pi epsilon not times the log of l plus the square root of z squared plus l squared divided by l by minus l plus the square root of z squared plus l squared.
02:21
Squared which we can then simplify to lambda over 2 pi epsilon not times the natural log of l plus the square root of z squared plus l squared all over z next for the last charge distribution c we have the potential v as 1 over 4 pi epsilon 0 times the integral from 0 to r of sigma times 2 pi r d r divided by the square root of r squared plus z squared this is equal to 1 over 4 pi epsilon not times 2 pi sigma into the square root of r squared plus z squared and we evaluate this between 0 and capital r, which gives us sigma over 2 pi epsilon 0 into the square root of r squared plus z squared minus z.
04:00
Now that we have the potentials, we can find the electric fields.
04:05
And we'll start with a, the electric field e, is equal to minus 1 over 4 pi epsilon 0 times 2 q times minus a half times 2 z over z squared plus d over 2 all squared all to the power 3 over 2 z hat which we can then simplify to 1 over 4 pi epsilon 0 times 2 qz over z squared plus d over 2 all squared to the power 3 over 2 z hat...